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For sure, $\ell^2$ is larger than $\ell^1$, because for $|x|<1$, $|x|^2<|x|,$ that is, $||x||_2\leq||x||_1.$

But using Cauchy-Schwarz inequality, I get a "wrong" comparison:

$$||x||_1=\sum_i|x_i|\leq\left(\sum_i|x_i|^2\right)^{\frac{1}{2}}\left(\sum_i 1\right)^{\frac{1}{2}}=\left(\sum_i|x_i|^2\right)^{\frac{1}{2}}=||x||_2.$$

What is going wrong here? Thanks.

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What two vectors are you applying the inequality to? For the vectors $(x_i)$ and $(y_i)$, you would know $\bigl(\sum x_iy_i\bigr)^2\le\sum x_i^2\,\sum y_i^2$. If $(x_i)=(y_i)$, then you obtain an identity. – David Mitra Oct 6 '12 at 14:32
@DavidMitra Another vector is $1$... See my edited formula. – Vladimir Oct 6 '12 at 15:22
@JackWitt Where do you take the sum? – Davide Giraudo Oct 6 '12 at 15:37
up vote 3 down vote accepted

What is wrong is that you are saying that $\sum_i 1 =1$, which is only true when $i$ runs over a set of exactly one element.

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So... Otherwise this sum is meaningless, right? – Vladimir Oct 6 '12 at 18:19
Yes. The Cauchy-Schwarz inequality applies to two sequences in $\ell^2$. The sequence $(1,1,\ldots)$ is not in $\ell^2$. If you try it with $n$-tuples instead of sequences, you get the estimate in the first sentence of Davide's answer. – Martin Argerami Oct 6 '12 at 19:30
Thank you very much! – Vladimir Oct 6 '12 at 19:34
You are welcome. – Martin Argerami Oct 6 '12 at 20:37

Cauchy-Schwarz inequality gives $\sum_{i=1}^n|x_i|\leq \sqrt n\sum_{i=1}^n|x_i|^2$. Actually, taking the sequence $x^{(n)}=(\underbrace{1,\dots,1}_{n\mbox{ times}},0,\dots)$, we get $\lVert x^{(n)}\rVert_1=n$ and $\lVert x^{(n)}\rVert_2=\sqrt n$.

But we indeed have $\lVert x\rVert_2\leq \lVert x\rVert_1$ as $$\lVert x\rVert_2^2=\sum_{j=0}^{+\infty}|x_j|^2\leq\left(\sum_{j=0}^{+\infty}|x_j|\right)^2.$$

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